Question: A person stands $30$ meters east of an intersection and watches a car driving away from the intersection to the north at $17$ meters per second. At a certain instant, the car is $16$ meters from the intersection. What is the rate of change of the distance between the car and the person at that instant (in meters per second)? Choose 1 answer: Choose 1 answer: (Choice A) A $8$ (Choice B) B $34$ (Choice C) C $\sqrt{1189}$ (Choice D) D $36.125$
Solution: Setting up the math Let... $a(t)$ denote the distance between the car and the intersection at time $t$, $b$ denote the distance between the person and the intersection (which is always $30$ meters), and $c(t)$ denote the distance between the car and the person at time $t$. $a(t)$ $b$ $c(t)$ We are given that $a'(t)=17$ and $b=30$. We are also given that $a(t_0)=16$ for a specific time $t_0$. We want to find $c'(t_0)$. Relating the measures The measures relate to each other through the Pythagorean theorem: $\begin{aligned} \,[a(t)]^2+b^2&=[c(t)]^2 \\\\\\ [a(t)]^2+(30)^2&=[c(t)]^2 \end{aligned}$ We can differentiate both sides to find an expression for $c'(t)$ : $c'(t)=\dfrac{a(t)a'(t)}{c(t)}$ Using the information to solve In order to find $c'(t_0)$ we need to find $c(t_0)$. Using the Pythagorean theorem and the fact that $a(t_0)=16$ and $b=30$, we can find that $c(t_0)=34$. Now let's plug ${a(t_0)}={16}$, ${a'(t_0)}={17}$, and ${c(t_0)}={34}$ into the expression for $c'(t_0)$ : $\begin{aligned} c'(t_0)&=\dfrac{{a(t_0)}{a'(t_0)}}{{c(t_0)}} \\\\ &=\dfrac{({16})({17})}{({34})} \\\\ &=8 \end{aligned}$ In conclusion, the rate of change of the distance between the car and the person at that instant is $8$ meters per second. Since the rate of change is positive, we know that the distance is increasing.